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What does the balance ball actually indicate?
Given answers and comments from a previous question, I realize that I need to clarify what the ball actually measure. I think it does not measure side slip but sideways acceleration, but I'm not quite sure.
Let's say, I put a ball in a roller coaster. Whatever the maneuver performed, there can't be side slip (no wind, the roller coaster is basically a train). Those maneuver include:

boat turns (the ball should be on the external side of the turn)
straight horizontal trajectories (the ball should be centered)
straight trajectory with a bank angle (the ball should be on the lower side)
a simple roll

How does this differ from in flight situations? Given what the ball actually measures, what makes it reliable to be used as a side slip indicator?
bonus question: if I take a long aircraft (e.g. a AN-225) and put a ball in several places in the aircraft, will all those indicator concur in a not coordinated turn?

Suggest you add to last paragraph "And if one of those indicators is centered (indicating a coordinated turn, will they all be?"

I would highly recommend that you get an orientation flight with a local flight school to see how this instrument actually works in flight.  I think it would clear up a lot for you.

@MichaelHall It is scheduled... as soon as the covid19 confinement ends in my country

1. Preamble
The ball reacts to the body force (or acceleration), with the exact same working principle as a bubble level. In airplane frame, the force is expressed as:
$$Y+mg\cos{\theta}\sin{\phi}=m(\dot{v}+ru-pw)$$
where $Y$ is all external side forces minus gravitational component, $m$ is airplane mass, $g$ is gravitational acceleration, $\theta$ is pitch angle, $\phi$ is roll angle, $\begin{bmatrix}u & v & w\end{bmatrix}^T$ is the inertial velocity in vehicle frame, $\begin{bmatrix}p & q & r\end{bmatrix}^T$ is the angular velocity in vehicle frame.
First, as any accelerometer, the ball does not react to gravity. Second, and this is a preference in bookkeeping, you can consider the accelerometer (or the ball) to only react to the external forces (i.e. $Y$), or only react to the inertial forces minus gravity (i.e. the right side of the equation minus the gravitational component).
2. In Flight
Taking thrust out of the question, then in flight, the only thing that can contribute to $Y$ is aerodynamic forces. What's the largest source of lateral force in air? Sideslip. There you have the relation between ball measurement and sideslip angle measurement. If you're unconvinced, read the sub-sections below.
2.1 Symmetric Thrust and Airframe
For symmetric thrust and airframe, the only yaw moment sources come from aerodynamics, and the largest contributors are sideslip and rudder (ailerons and spoilers also contribute due to differential drag, but the effects are even smaller).
Using the B737 model in AVL at Mach 0, flap 0 and AOA 0 for a rough idea, -5 deg rudder generates about 4.4deg of sideslip (nose left), at which point their yaw moments cancel each other out. However, the total $C_y$ is 0.055 to the right.
Comparing the stability/control derivatives (per rad), $C_{y_\beta}$ is -1.2, whereas $C_{y_{\delta_r}}$ is 0.44, which means that at the aforementioned conditions, the sideslip is overpowering the rudder by 140%, in terms of side force.
2.2 Asymmetric Thrust
In the case of one engine inoperative (OEI), we need to overcome the yaw moment from thrust asymmetry via rudder.
For wings level, there is side force from rudder that must be compensated. This can be accomplished by generating more sideslip into the live engine (which means more rudder deflection is needed than to overcome the thrust yaw moment). At steady-state, the net side force is zero, the ball is centered, but there is steady non-zero sideslip.
If we allow the wings to bank toward the live engine, a portion of gravity can be used to compensate for the side force from rudder, which means that less sideslip induced side force is required to maintain steady-state (therefore less rudder required). However, since the total side force minus gravity ($Y$) is non-zero, the ball is skewed to the live engine.
In either case, with thrust asymmetry, the relationship between sideslip and side force are "broken".
3. On Ground
On ground, any point of contact with the ground provides additional sources to $Y$. This is why tilting the bubble level with your hand makes it not measure level: it's not that the level is responding to gravity, but rather that it's responding to the normal forces exacted by your hand.
For an airplane, the source is obviously coming from the landing gear contact. At high speed taxi on a straight runway, with crosswind, the sideslip creates aerodynamic side force. The landing gear resists lateral motion via an opposite side force (however, you'd need to deflect nose wheel or rudder to resist yaw moment). At steady-state, the yaw string would measure sideslip, but the net side force is zero, so the ball would measure nothing.
Things are different when it's a circular runway. Because the airplane needs to track a curvature, the combined lateral forces must provide the centripetal forces necessary to maintain the track curvature. Therefore, the ball would react to the equivalent centrifugal force.
4. Sensor Positioning
Thus far, we've been assuming that the ball is placed exactly at the airplane CG. If the sensor is offset, then it will also react to the angular rates. That's why large airplanes need to calibrate the sensor measurements to the assumed CG position.
To answer your last question, unless the airplane is tracking linear motion, placing the sensor at different locations will give you different readings.

Regarding "Sensor Positioning"-- am I wrong (second-to-last paragraph of my answer) to say that the instruments will read the same if the rotation rates are constant? Can you help me understand why? Seems to me that steady pitch and yaw rotations would make an apparent centrifugal force only in the longitudinal direction which should not affect the inclinometers.

@quietflyer For a uniform angular velocity, positioning of the sensors will distort the measurements (cross product arm). It will be a longitudinal bias only if the cross-product is exactly zero in the y-axis, which is most often not the case.

you speak of "the largest source of lateral force in air". How large is it compared to other sources?

@ManuH We are referring to all engines operating (or no engine operating), and in flight again, correct? OEI is a different scenario, and on ground is a much larger departure (as noted above).

@JZYL I was more thinking in a glider, but whatever the aircraft, its flight is nominal, yes. Yet, in your paragraph, you speak of the "largest source of lateral force in the air", thus it means other sources exists and are less powerful. How less?

@ManuH Edited to show some numerical comparison

@ManuH the rudder itself can be a source of lateral aerodynamic force even if the fuselage is perfectly aligned with the airflow (i.e. sideslip is zero) -- that's what the example of the twin-engined aircraft with one failed engine is all about.

It's not aerodynamic force, It's aerodynamic acceleration. Look at your own equation. Y + gravity force.  If it is not accelerating there is no aerodynamic contribution to G forces.  The rudder is the source of aerodynamic acceleration when the curved tube is positioned with its ends to the sides and the bottom towards the floor in an aircraft.  The ball simply rolls to the "lowest" part of the curve, which is the combined G force of gravity and aerodynamic acceleration.  Note if ends were fore and aft, it would measure acceleration and deceleration of airspeed.

@RobertDiGiovanni What will an accelerometer measure if it's in free-fall?

@JZYL thanks for asking, and to all, thanks for continuing the interesting discussion.  In a free fall, in atmosphere, the object will reach terminal velocity, the ball will sit at the bottom of the  u shaped tube.  The drag vector is up, gravity down.  Now lets have a gust of wind push it from the side.  The tube will have lateral acceleration, and the ball will move in the opposite direction by G force concept but in reality, the tube is accelerated sideways by the gust and the ball is "left behind".   Comprende?

@JZYL an object in free-fall (in atmosphere) is closest to "weightless" when you first drop it.  Gravity vector down, ever increasing drag vector up.  A plot of acceleration over time will show a decrease from 9.8 $m/s^2$ to 0 $m/s^2$.  The only way to make the ball move "up" in the tube is to provide additional down thrust, and to minimize drag.  Sound familiar?

@RobertDiGiovanni, “free-fall” is a precisely defined physical term that is incompatible with being in atmosphere. In free-fall, accelerometer reads zero, by definition. In atmosphere, the fall will be opposed by drag (so it is not free). The plot of acceleration as derived from position over time in the Earth reference frame will show a decrease from 9.8 m/s² to 0. But an accelerometer measures relative to free-fall, so it will measure a decrease from 0 to -9.8 m/s² (if I keep positive pointing down). This -9.8 m/s² is the aerodynamic force (vector sum of all) divided by mass of the object.

Regarding “external forces minus gravitational component”, I generally prefer to consider gravity an inertial force as per general relativity, because then inertial forces in the body frame exactly oppose external forces on the body. But in Newtonian physics gravity was considered external force, so then it is needed, and it is correct either way.

@Jan Hudec to hear from you.  I prefer 9.8 $m/s^2$ to 0 $m/s^2$ for our book keeping, as it is the aerodynamic part of JZYLs equation, and at steady state fall, 0 represents the aerodynamic G force contribution, there for 1 person in the elevator weighs 1G from gravity alone. - 9.8 would work in the initial state, as subtracted from gravity to give 0 G (the instant the cable broke).  The confusion here seems to be about forces, acceleration (which results from unbalanced forces), and aerodynamic "G force" which I call "inertial resistance".

@RobertDiGiovanni, yes, we can stick to the Earth frame of reference and Newtonian understanding of gravity. At steady state, the aerodynamic force, $Y$, is ‘1G’ (=$mg$) upward as it balances gravity. There is no acceleration in the Earth frame of reference, but accelerometer measures relative to free-fall, so it indicates $1g$ up, corresponding exactly to the $Y$. The part about including/excluding gravity is confusing, because gravity is excluded from external forces, but included in the inertial forces. The equation is certainly correct.

@RobertDiGiovanni, defining additional terms like ‘aerodynamic acceleration’ and ‘inertial resistance’ is counterproductive here because the accelerometer reading is simply sum of the aerodynamic forces divided by mass (and then we limit ourselves to just the lateral component in body frame, but that's the next step).

A note regarding terminology: “gravity” does not mean acceleration due to gravitation, it really does mean the acceleration due to inertial forces including gravitation. A more appropriate name would be “weight acceleration”, because it is what creates weight. By default it means the gravity in the reference frame of Earth surface, but note that there it includes the centrifugal force from Earth rotation and excludes gravitation from Sun, because that is balanced by orbital centrifugal force. In this sense, accelerometer measures local gravity and only that.

This is a great discussion! Can someone explain why "the ball does not react to gravity" please (the link is broken)? I can't make sense of this statement (unless the ball is in a fluid where the buoyancy of the ball is equal to the weight of the ball). Thanks!

When you see inclinometers placed at different locations like that it's for measuring level on the ground for fuel and center of gravity measurement or other leveling purposes.
The one on the instrument panel works for measuring skids in flight because when you are skidding you are moving in an arc, or turning, to one degree or another, because the fuselage presented to the side creates lateral force, plus the thrust line offset also produces a lateral force.  The ball tells you that you are in a flat turn in other words, like you're in a car turning a corner. If no lateral movement occurred while skidding and there were no other accelerations, the ball would sit in the middle.
As an illustration, you can see this when you do engine failure training in a piston twin.  When you are single engine, you have rudder input to stop the yaw caused by the offset thrust line of the live engine.  The rudder's side force offset's the engine's thrust line from straight forward to angled several degrees in the direction of the rudder side force.  You end up flying slewed toward the dead engine, a very draggy condition because the airflow is not aligned with the fuselage, but you aren't actually turning, as in changing heading, and the ball sits in the middle even though the plane is technically in a side slip.  If you are in a marginally powered piston twin on a hot day at gross weight at 5000 ft, this slewing inefficiency can make the difference between climbing at all, or not.
To fix that, you lower the wing into the live engine, about 5 deg of bank.  This induces a side slip component that offsets the rudder's lateral thrust component.  You find yourself in a strange condition: you are banked to the left 5 degrees, but flying straight through the air, with the skid ball offset toward the bank by about half to one ball width. It's not centered in the vial, but it's "centered" as far as indicating no lateral acceleration, that is, it's pointing down at the center of the earth.
One huge advantage of yaw strings on the windshield, as used on helicopters, gliders, and jets (and really can be used on any airplane without an engine in the nose), is they indicate true sideslip directly and don't depend on lateral acceleration.  In the multi-engine scenario I described above, the ball will be centered with no bank into the live engine, but the yaw string will reveal what is actually going on and will be offset to show the slewing flight; then when you bank 5 deg into the live engine, the ball will be offset but the yaw string will be straight.

In a forward slip you are not moving in a circle, because you are compensating the side-force by banking, and the ball is offset as corresponding to the slip (it's the same as the engine out, but this time it is an actual slip).

@Jan Hudec of all your comments, this one reaches the source of the confusion the best.  The ball, In this case, with no lateral acceleration, simply lies at the lowest point in the tube, closest to earth, not because the plane is slipping, only because it is banked.  Even if the plane flew completely sideways, in steady state, the ball would be in the same spot because only gravity is acting on it. Well done!

Well that's exactly what the caption beside the skid ball in my illustration says.

Imagine you have a rope with a heavy weight attached to it dangling under the aircraft. In straight and level flight, this weight will hang directly under the aircraft. As the aircraft turns, it banks and the ball swings to the outside of the turn, but it is still straight down (referencing the body of the aircraft). If the aircraft were to initiate a turn without being coordinated, the weight would swing outside (skid) or inside (slip).
This is why it's called a "slip/skid" indicator. The ball tells you where "g" is (not gravity, but the direction of force on the aircraft). If you were to roll the aircraft upside down, the ball would fall either left or right rather than remain in the middle (if you do a 1-G full-roll though, the ball should remain centered as your/the aircraft's sensation of "down" is constant).
And yes, if you put a slip/skid indicator around certain parts of an aircraft they should indicate the same thing. This is because an aircraft is a rigid-body system (for the most part), all parts of the aircraft should experience the same forces in flight as the other parts. This isn't strictly true since things like wing-tips flex in flight, sometimes significantly, but you can think of it that way.

Ron, since we're all being really picky here, the only way the ball stays centered in a roll would be far enough away from the CG with a roll fast enough to "pin" it to the bottom of the curved tube.  You may be thinking of Bob Hoover's famous barrel roll.  "If you were to roll the aircraft...the ball would fall left" see John's diagram.  Now, if you stopped the roll at 20 degrees and put in left rudder, the combination of gravity, and lateral acceleration (to the left) which creates a "G or centrifugal" force to the right,  recenters the ball.

Much of this answer is based on the related answer How does sideslip indicator react during crosswind?.
We should start by understanding what the deflection of the "inclinometer" or "ball" actually indicates.  To a good approximation, it responds to the net sum of all lateral (sideways) accelerations acting on an aircraft, excluding the lateral acceleration component due to gravity.  By "lateral" or "sideways" we mean in the aircraft's own reference frame -- for example, in a turn, the direction meant by "lateral" or "sideways" would be constantly changing as seen by an observer on the ground, or for that matter, an observer in a balloon drifting by with the airmass.  In a turn, by "lateral" we don't mean the "centripetal" direction-- rather we specifically mean "sideways toward the wall of the cockpit".  In other words, the direction that is parallel to the aircraft's wingspan.  In a coordinated turn, there is definitely a lateral acceleration component, but the aerodynamic force in the lateral direction is zero.  In other words, in a coordinated turn, the net lateral acceleration other than that due to gravity is zero.  To a first approximation, the only thing that causes an aircraft to experience a lateral acceleration, other than gravity, is the aerodynamic sideforce generated by a sideslip as the airflow strikes the side of the fuselage.  That is why the ball serves as a slip-skid indicator, i.e. a sideslip indicator.***  There are some reasons why the ball is not a perfect indicator of sideslip, which will become evident later in this answer, but the idea that the ball is basically a sideslip indicator is a good starting point.
To be complete we should also point out that for a given aerodynamic sideforce, the motion of the ball is also influenced by the G-loading, or more precisely, by the upward aerodynamic force generated by the wing.  The higher the "upward" G-loading, the more aerodynamic sideforce it takes to cause a given deflection of the ball.  At zero "upward" G's, the slightest aerodynamic sideforce will put the ball all the way into one of the far corners of the glass tube, and at negative "upward" G's, the ball tends to just stay stuck in one of the corners even when sideforce is zero.  This is why many aerobatic aircraft have a second slip-skid ball that is mounted upside-down.
If the ball is an accelerometer, how can it be deflected in a situation like a steady-state wing-down slip along a straight-line flight path, where net acceleration is clearly zero?  The answer is that net acceleration is zero, but there is still an aerodynamic force component, and therefore an aerodynamically-driven acceleration component, acting in the sideways direction in the aircraft's reference frame, caused by the air impacting the side of the fuselage.  Just as a panel-mounted G-meter reads "1" not "0" in straight-and-level flight-- net acceleration is zero, but there is still an aerodynamic force component acting in the upwards direction in the aircraft's reference frame-- the lift force from the wing.  Just as the slip-skid ball measures the net lateral (sideways) acceleration other than that caused by gravity, so too does the G-meter measure the net vertical acceleration component other than that caused by gravity, which is essentially just the upward acceleration caused by the wing's lift vector.
What reference frame is the slip-skid ball operating in?  We've seen that the slip-skid ball measures lateral acceleration (other than that due to gravity).  We can measure net acceleration in any valid inertial reference frame and we'll get the same answer, except for differences due to the tilt of one frame relative to the other.  An accelerating airmass is not a valid inertial reference frame.  Nor is the aircraft itself, unless net acceleration is zero.  Therefore it is not wrong to assert that in the context of an accelerating airmass, the slip-skid ball is operating in the reference frame of the earth, rather than the reference frame of the aircraft or the airmass, except that the reference frame is tilted to match the orientation of the aircraft in space at any given instant in time, and so is constantly changing its orientation with respect to the orientation of the earth's reference frame.  In other words the direction called "lateral" is fixed with respect to the aircraft, not the earth-- it changes (relative to the earth) as the aircraft changes heading, or as the aircraft banks.  Yet the actual acceleration in that (ever-changing) lateral direction is measured in relation to a valid inertial reference frame, not in relation to the aircraft itself.  It is a valid question to ask how it can be that the slip-skid ball which in a sense is tied to the earth reference frame, can give essentially the same indication as the yaw string.  Perhaps the above paragraphs have shed some light on those questions-- the answer has to do with the fact that the slip-skid ball is only measuring the lateral component of acceleration, and the fact that the lateral component of acceleration is intimately related to sideslip.  Keep in mind that in many instances in this answer, we use phrases like "in the aircraft's own reference frame", but we really only mean to describe the orientation of the reference frame involved.  At any instant time, the actual reference frame that an on-board accelerometer such as the G-meter or the slip-skid ball is operating in would indeed be an actual valid inertial reference frame such as that of the earth itself, just tilted differently.  It is correct to observe that an accelerometer is not actually operating in the reference frame of the aircraft itself in any case where the aircraft is accelerating, or in the reference frame of the airmass in any case where the wind is not steady.  If the slip-skid ball were actually operating in the reference frame of the aircraft itself, it would always read zero, and likewise the G-meter.
In any situation where lateral accelerations can be caused by something other than sideslip, the ball is not perfect as a slip-skid indicator.  One example would be any case where the fuselage is streamlined to the relative wind but the rudder is deflected.  The sideforce from the rudder itself will move the ball slightly off center opposite to the direction that the pilot would normally expect, i.e. in the same direction as the rudder is deflected.  Therefore when using the rudder to streamline the fuselage to the airflow, it is optimum to leave the ball slightly off-center in the direction that it would be deflected if the rudder were not applied at all.  Normally this effect is negligible in actual flight, but not always.  The scenario of a twin-engine airplane with one failed engine is the best-known case where a pilot will typically actually take this effect into account, leaving the ball significantly deflected in the direction of the pilot's rudder input, i.e. toward the working engine  -- see the eighteenth paragraph, and also the footnote, in  this related answer for more on this.  Of course the sideforce from the rudder will cause a turn unless we cancel it by banking toward the good engine.  But fundamentally, the bank is not what is displacing the ball-- for the same rudder input, the ball would show the same displacement regardless of whether the plane was banked as needed to continue in a straight line, or was kept wings-level and allowed to do a flat turn toward the good engine.  The yaw string would be centered in either case.  (Obviously, if we constrain the flight path to be linear, then everything changes and the slip-skid ball doubles as a bank angle gauge.  From this viewpoint, we know we should bank a bit toward the good engine, but the reason why is not obvious.)
The tail rotor of a helicopter also generates an aerodynamic sideforce.  Again, just as with the twin-engine airplane with one failed engine, when the fuselage is streamlined to the airflow in forward flight, the slip-skid ball will be slightly off-center, deflected opposite to the direction of the tail rotor's sideforce-- at least in cases where the body of the helicopter tends to tilt along with the disk of the main rotor.  If the disk of the main rotor can be tilted to counter the tail rotor's sideforce while the fuselage or body stays completely vertical, then the slip-skid ball can stay centered even when the fuselage or body is completely streamlined to the airflow.
A little-known fact is that it is also possible for the yaw string and slip-skid ball to disagree when an aircraft (glider) is being towed, or when an aircraft (airplane) is towing another aircraft, because the towline can impart a lateral force component whenever the two aircraft are not directly behind each other. In a turn on tow, where the heading of each aircraft is tangent to the circular path of turn, the towline is imparting a lateral force toward the center of the turn on each aircraft, so the ball will be displaced toward the outside when the yaw string is centered.
Note that we will see less deflection of the slip-skid ball for a given deflection of the yaw string when a gust strikes a heavily loaded aircraft, than when it strikes an aircraft of identical shape and size but less mass, travelling at the same airspeed.  A much simpler analogy would be two toy cars of the same shape and size but different mass, each with a flag on it. When a wind gust strikes both cars and blows both flags in an identical manner, the wind will accelerate the lighter car faster than the heavier car. The flags are analogous to the yaw string, and the cars are analogous to the aircraft accelerating sideways and tending to "leave the ball behind", so that the ball deflects sideways inside its tube.
Likewise if the pilot uses the rudder to induce a given sideslip angle as measured with the yaw string, the ball will deflect less when an aircraft is heavily loaded than when it is lightly loaded, if the shape and size and speed of travel are all identical.  This means that the heavily-loaded aircraft will require less bank angle to travel in a straight line while the pilot is using the rudder to maintain a given sideslip angle, as measured with the yaw string.  Or in other words, for a given aircraft travelling at a given airspeed, a given deflection of the ball indicates a larger sideslip angle when the aircraft is heavy than when the aircraft is light.  However this is not something that ever needs to be considered in actual for flight for any practical reason-- normally we just keep the ball centered, unless we are using the wing-down (slipping) method of crosswind correction just before touch-down, in which case we bank as needed to allow the desired ground track while keeping the nose aligned with the runway heading.
Obviously on the ground the connection between sideways airflow (sideslip) and lateral acceleration is broken, unless the aircraft is wings-level on a flat icy surface.  The wind can blow (creating sideslip) without accelerating the aircraft.  Also unless the ground is level, it can push up against the tires in a way that exerts a lateral force component on the aircraft.  (Even though the ball doesn't sense the downward acceleration component due to gravity, it does sense the upward accceleration component of the ground pushing up on the tires-- this is why the ball functions as a bubble level when the aircraft is parked and stationary, just as it does in flight whenever net acceleration including gravity is zero.)  The question also asked about a roller coaster or train -- there are many situations where the track will exert a lateral force on the vehicle, so a slip-skid ball would not always be centered.  Clearly sideslip (sideways wind) has no influence on the slip-skid ball at all in this situation, just as in the case of the aircraft on the ground.
The ball is not quite perfect as a sideways accelerometer.  Changes in the rate of yaw rotation about the aircraft's CG can also influence the ball.  Imagine that we mounted the aircraft on a pivot at its CG in a vacuum-filled hangar and twirled the airplane like a pinwheel in the yaw dimension.  Whenever the yaw rotation rate was increasing, the ball would deflect opposite the direction that the nose was moving.  The further the cockpit from the CG, the more pronounced this effect would be.
In other words, a change in yaw rotation rate causes a local lateral acceleration, as measured at some specific point along the length of the fuselage.  Also, changes in the rate of pitch rotation change the local G-loading as felt at given points along the length of the fuselage, which changes the indication of the slip-skid ball-- see below for more.

bonus question: if I take a long aircraft (e.g. a AN-225) and put a
  ball in several places in the aircraft, will all those indicator
  concur in a not coordinated turn?

In a steady-state turn-- meaning constant rates of rotation in pitch and yaw-- then the only apparent centrifugal force induced by the rotation will act in the longitudinal direction, and all the inclinometers will concur, regardless of whether they are centered or not.  (Extra detail-- strictly speaking, for this to be true, the inclinometers must all be mounted along the aircraft centerline, so that they are exactly tangent to the arc of movement during a yaw rotation.) Whenever the yaw rotation rate is increasing, inclinometers at the front will be biased toward a deflection toward the outside of the turn and inclinometers at the rear will be biased toward a deflection toward the inside of the turn.  Whenever the pitch rotation rate is increasing, the "felt" G-loading will be increased at the front of the aircraft and decreased at the rear of the aircraft, which will tend to bias the inclinometers at the front of the aircraft toward less deflection (in whatever direction they are deflected), and will tend to bias the inclinometers at the rear of the aircraft toward more deflection (in whatever direction they are deflected).  Note that as we progressively increase the bank angle, the yaw rotation rate increases and then starts to decrease again, while the pitch rotation rate continually increases.  A shallow-banked turn involves mostly yaw rotation, while a steep-banked turn involves mostly pitch rotation.  So we can think through how the indications of the inclinometers located at various points along the fuselage will change over time as we progressively increase the bank angle.  In short, all the indicators will not always show exactly the same reading.
Normally these effects are negligible in actual flight.
Note that yaw strings mounted at various positions along the fuselage will not show exactly the same deflection during a turn-- the rotation of the aircraft in the yaw axis causes different parts of the aircraft to be moving through the airmass in different directions at any given instant in time.  In other words the undisturbed free-stream airflow or "relative wind" is actually curved in a turn. To read about how one pilot takes this into account in actual practice, see the Soaring magazine article Circling the Holghaus way by Richard H. Johnson.  Links to some other articles discussing the curvature of the relative wind in turning flight are embedded in this ASE answer.
If you really want to challenge yourself to better understand the behavior of the slip-skid ball, consider the following thought experiments: A) While flying in a moderate to steep turn, the pilot pushes the stick forward hard to unload the wing, or pulls the stick aft hard to load up the wing with extra G's.  Obviously the turn rate will vary.  Does the slip-skid ball move very much?  Does it move at all?  Does the yaw string (if present) deflect?  B) Imagine a wingover-like maneuver where the aircraft is "unloaded" to less than 1 G while steeply banked, in a (initially) near-level or slightly-climbing pitch attitude.  Obviously gravity is accelerating the flight path downwards, and obviously this downward acceleration is largely in the lateral direction in the aircraft's reference frame, at least initially.  If the pilot uses the rudder as needed to "help" the vertical fin keep the nose pointing directly into the airflow so that the yaw string stays exactly centered, is the slip-skid ball deflected?  Now go do the experiments, ideally in a sailplane or some other aircraft with a yaw string and no propwash over the nose.  The latter experiment in particular cannot be done in a useful way with no yaw string-- the aircraft definitely will sideslip if the pilot doesn't use the rudder as needed to prevent it.

"the only thing that causes an aircraft to experience a lateral acceleration, other than gravity, is the aerodynamic sideforce generated by a sideslip as the airflow strikes the side of the fuselage."  Are you saying that in a coordinated 6G level there is no acceleration?  And what do you mean by lateral acceleration due to gravity?  Gravity only acts vertically.

@MichaelHall  In the aircraft's reference frame, in a coordinated turn, you have the lift vector acting in the "upward" direction (is there some better word for this than "upward"?), and you have thrust and drag acting in the longitudinal direction, and no aerodynamic force acting in the lateral direction.  Meanwhile gravity does have a component in the lateral direction.  From a reference frame tilted to match the aircraft's orientation in space at any given instant.  Which is the reference frame that the slip-skid ball lives in.  Not from the ground reference frame.

@MichaelHall the reference frame I'm using is established in second sentence of second paragraph. But now added two more following sentences to make more clear.

@MichaelHall would some other word be more intuitive for the direction I am talking about?  I used to say "spanwise" but then people tend to imagine one is speaking of airflow moving from the root to each wingtip, which has nothing to do with it.

@MichaelHall remember that the skid ball sums gravity + lateral acceleration, It's a resultant.  When banked at the right angle, the resultant is aligned with the aircraft's vertical axis and the ball is centered in its vertical axis reference index, the two lines.

@JohnK -- what would be the meaning of being banked at the "wrong" angle?  Given the specific constraint of a steady-state turn (all rotation rates constant) I guess we can say it's possible to be banked at the "wrong" angle for the turn rate-- but to a good approx., this can only happen because we are dragging the side of the fuselage sideways through the air.  Since sudden pitch inputs (sudden changes in a-of-a) change the G-load and the turn rate but don't affect the ball (much), I tend to avoid explanations that say the ball tells you whether the turn rate is "right" for the bank angle.

By "right" meant when the angle of bank and the rate of turn are aligned so that the resultant "gravity" is aligned with the airplane's vertical axis.

@quiet flyer, I follow you now.  We were just looking at things differently.  FYI, one of the downvotes was mine.  Same criticism I have levied before, you are obviously very knowledgeable but your answers to simple questions are overwrought.  It can be excruciating to sift through everything you are saying to distill out the applicable nuggets of wisdom.  Ron Beyer's answer above is perfect: On-point, with no extra information to obfuscate the base question.  (reference my comment on your brilliant paragraph 8 on another extremely lengthy answer...)

Your steady-turn paragraph isn't right.

@MichaelHall -- I guess the key point re "lateral" acceleration is that in a coordinated turn, the NET force has a lateral component but the net AERODYNAMIC force does not.

@JZYL -- are you talking about the paragraph immediately under where I quote the "bonus question"?  If so, I know you said that already in another comment in response to my question to you but I don't see why.  If hold an inclinometer in my hand and whirl it around in the directions that simulate a pitch or yaw rotation at a constant rate, I don't think it will deflect.

@JZYL -- or if you meant a different section, please say which, and what is wrong with it.

Now grab a piece of board. Attach the level offset from the center both longitudinally and laterally. Now spin it.

Ah, I see, the problem is that the instrument is not mounted tangent to the arc of rotation.  I'm having trouble imagining that be a significant source of error in the specific case where we have mounted inclinometers along the length of the fuselage.  Are you essentially saying that the inclinometers need to be along the aircraft centerline?

For small airplanes, it's located as close to the CG as possible. There are acceptable error tol.

Re "At any instant time, the actual reference frame that an on-board accelerometer such as the G-meter or the slip-skid ball is operating in would indeed be an actual valid inertial reference frame such as that of the earth itself, just tilted differently"-- should be "At any instant in time..."


text:
How do flight procedure designers decide when to assign an altitude constraint to a waypoint?
I am studying procedure design manuals (TERPS, PBN 8260.58A manual, ICAO PANS-OPS 8168 manual etc.) and I haven't found any guidelines concerning altitude constraints that can be imposed on a waypoint.
Take, for example, JFK's DEEZ FIVE departure. CESID has an "AT or ABOVE 2500" altitude constraint. Is there a reason that an altitude constraint was imposed on CESID, rather than on e.g. SKORR or YNKEE?
Another example, as far as STARs are concerned, is SEA's HAWKZ SEVEN. There is a Minimum Enroute Altitude for each leg, which is more than enough both for obstacle clearance and for communication. Why did the designers decide to an "At or ABOVE FL220" at PTERA? And why at PTERA and not e.g. at KRIEG, KNGDM or WRUSL?
Can someone point me in the right direction?

Altitude limits on these procedures have to take many different things into account, including obstacle clearance and communication, but also radar coverage, separation from other procedures and airspace, the overall profile of the procedure, noise abatement, and more.
The at or above 2500 restriction on the DEEZZ is similarly included in the other SIDs at KJFK. The main reason could be that the NYPD operates helicopters out of Floyd Bennett Field below that area, and the Jamaica Bay seaplane base is also nearby. Another reason could be to keep aircraft above the approach for runways 4L and 4R, though I'm not sure they would use those at the same time. It could also help noise abatement for the homes and parks in the area.
In the case of the HAWKZ, my guess on the altitude restriction is to keep all the aircraft on those segments at approximately the same altitudes. It seems like an aircraft could be lower than FL220 at PTERA without other issues, but the lower an aircraft descends, the slower it flies, and it would become difficult to work a slower aircraft in with the faster stream of traffic at the higher altitude. This sort of restriction is not uncommon on STARs that begin relatively far out.

Keep in mind McChord AFB is just to the South of SEA, and that they have published high altitude instrument approach procedures that might drive keeping commercial traffic higher than terrain would otherwise dictate.

@MichaelHall KTCM is a good point, but PTERA is closer to PDX than TCM, and I don't see even high altitude procedures for either one in that area.

Good point as well.  Still, it's 83 NM from SEA, and a roughly 265' per NM descent gradient isn't too steep.  I think the best we can do is speculate...

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Do the regulations allow DME from an expired IFR-certified GPS with an out of date database to be used for this ILS approach under an IFR plan?
Since we cannot fly an RNAV approach with out of date GPS database, the approach in question is KCRE ILS 23. Whether it was asking for V2F or the full approach starting from ASHES (which is 18.5 DME to CRE VOR). Are intersections "moved" all that often? Is there a way to verify if this happened recently?
As of this writing, the approach plate has been last amended June 22, 2017.

https://aeronav.faa.gov/d-tpp/2108/05097IL23.PDF

@ymb1, there are some good quotes from the regs in the answer here:  https://aviation.stackexchange.com/questions/11568/can-i-use-an-expired-ifr-certified-gps-with-an-out-of-date-database-in-lieu-of-d?rq=1

"an expired IFR-certified GPS [...] for an ILS approach?" Why would you need a GPS to fly an ILS approach? You rather need an CDI. Or do I  miss something??

Using an IFR certified GPS that has an out of date database is supposedly acceptable en route assuming the pilot has verified that the waypoints have not changed. In my particular case, the flight plan was KTTA -> FAY -> KCRE. FAY VOR hasn't moved last I heard. So, if one gets the visual for R23 or R5 at KCRE, should be good to go from my understanding. However, I was wondering whether using it for the ILS approach was acceptable in IMC? Let's assume that the fixes haven't moved (maybe that assumption is safe considering the last amended date is 4 years ago?)

@757toga: I see, the question would be more understandable as "can I use an out of date db GPS in lieu of the DME to locate a fix" then. Now I wonder if the DME is actually required for the ILS and LOC approaches or for the LOC only.

I made a minor edit to the title question to capture what GPS information is being used during the ILS, and deleted my earlier comments.

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text:
Is there a reason why the vertical lines of some airplane envelope graphs are not parallel?
I am working on a project to develop an easy solution for mass and balance generation for my aviation club. (First example below)
I checked the Graph of a PA-44-180 from the POH and wondered why piper chose to make the vertical lines non-parallel. (The lines diverge towards the top as shown in the second example below).
Is there any good reason to present the graph in such a way? Is there simply a mistake in the linear graph, or in the way I want to project it? Thanks in advance.
 vs

In my logic, piper decided to make the graph diverge towards the top, so that a pilot will be able to more precisely plot the actual calculated CG in the area, where more precision is required... that is my guess.

Not sure about that, since in the end the only thing that really matters is whether or not you cross the outer lines. What happens in between is not so important..

In my opinion, some older figures had an additional axis below, which was lost (maybe no more needed), but they keep the shape for consistency of older figures. (Possibly some GC-measure near wheels: more weight, more leverage, more deformation).

To continue on @GiacomoCatenazzi, the missing horizontal axis seems to be torque moment around the datum. By presenting it this way one can presumably more easily add contributions of additional weight, and retrieve the new location of the cog

Related, but not a duplicate: https://aviation.stackexchange.com/q/98806/7532  That question refers to the limitations themselves; this question is about how those limitations are presented. So not a dupe.

This kind of presentation is called a weight vs. moment 'fan chart'.  Although the labeled and drawn x-axis is in CG position, it is really calculated in terms of moment (inch-lb).  The reference datum is the location at the center of the chart with a vertical line.
Adding a mass at a location (relative to the datum) will cause a horizontal change -- independent of where you are vertically on the chart.  For example, filling a forward fuel tank from empty to full -- will always cause the same change in moment relative to a datum.  It will also cause the same vertical change.  So, you can compute this change once and then 'move it around' in terms of a sequence of changes.  This will tell you how the CG moves throughout a mission.
Any CG change will have an associated 'change vector'.  We can add these vectors head-to-tail to get the history of a CG during a load.

Notice in this example how the gear up / gear down lines are exactly the same horizontal width?  This is because raising and lowering the gear will always cause equal and opposite changes in moment.  The gear change vector stays the same.
Imagine if we extended the gear when the aircraft was at 30,000 lb (in this scenario).  We can quickly graphically compute the change in CG - to every point from there on down (ignoring that the gear would be separated from the airplane and other problems would occur if you put them down at that speed).

Very nice.  Someone please offer a bonus to reward this answer.

Would the change vectors work the same way on the graph the OP generated? Is there anything about the fanned-out vertical (ish) lines that makes the vectors work, that wouldn't work if those lines weren't drawn fanned out? As I read it, the fan itself is the central question, if it's essential, or just nice-to-have (and if the latter, why is it nice to have?)

If the OP plotted their chart in terms of Weight vs. Moment -- and then overlaid lines of constant CG, they would get a fan chart.  The fan is a bit of an illusion -- it is a consequence of doing the calculations.  Alternatively, if you were to plot constant moment lines on the OP's chart, it would form an upside-down fan.  Unfortunately, that isn't useful the way the traditional fan chart is.

Thanks @quietflyer


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What are the Garmin G-1000 LRUs exactly? Where are they located?
During my IFR flight training I was told to check the LRU status on the G1000 of the Cessna 172 I was flying. On the G1000 this is displayed as a list with many options showing a green tick or not. Now I am wondering how LRUs actually work and are integrated with the G1000 and their respective location within the aircraft?

Adding to @John answer.
LRUs for the Garmin 1000

What are the Garmin G-1000 LRUs exactly?

The LRUs, line replaceable units, are the discrete components which can be maintained and updated without returning to the manufacturer. The G-1000 Maintenance Manual describes all LRUs associated with the integrated avionics, e.g. the LRUs associated with instrumentation.

There are other LRUs: transponder, communication, navigation, etc, that cannot be detailed here, but the manual has a complete description and you can ask for specific details if necessary.
LRUs location

Where are they located?

LRUs in a large aircraft have dedicated electronic bays. On a Cessna 172, some LRUs have a front plate on the cockpit panel and the rest of the LRU is behind the panel. Remote LRUs are located at various places, often in the avionics shelf where they are air-cooled. The avionics shelf is behind the baggage compartment. Some LRUs are sensors, mounted at the most appropriate location:

Source: Civil Air Patrol
LRU status
LRUs status is indicated on the PFD and MFD, a red cross means the system doesn't receive the expected data fron a particular LRU:

See section 2.1 G1000 Alerting System in the manual for details by LRU. These status are also visible on the dedicated page:

LRUs are Line Replaceable Units. Parts that can be replaced "out on the flight line" without major disassembly.
In the case of the G1000, these are either black boxes themselves for different systems that interact with the G1000 and that it monitors, or in some cases components within the G1000's own electronics boxes that can be swapped out without removing the box itself (if not, then the box itself is the Line Replaceable Unit).
In such cases, these will be circuit cards that can be swapped out by opening a box, removing the card, and plugging in a serviceable one (like changing a RAM card on a personal computer).
This is a widespread practice in the avionics used in large aircraft. You try to get the lowest level of component you can to the LRU level to make repair as fast as possible as cheaply as possible while out on the line (usually at the gate).
What it means in practice is that when something goes wrong during operation, the component that is easiest and cheapest to replace on-the-spot, the LRU, is flagged so maintenance doesn't have to troubleshoot, just change out the part (of course, if that doesn't work, then the troubleshooting has to start in earnest).


text:
Can a bullet/finless rocket shape be a stable lifting body?
Consider a reentry vehicle that looks like this

If this re-entered the atmosphere from orbit, presupposing a sufficiently forward CG, could it stabilizer into a nose-first lifting attitude?
I'm thinking that as the AoA increases, the s shape will begin to behave like a canard planform, where the nosecone is the canard as it has a higher AoA (the bottom surface does, with the top surface being inactive), while the main body is the main wing.

Previous discussions have somewhat agreed that non-aviation aerodynamics questions are offtopic here. I can see how this one might be borderline so I'll not mod-hammer it just yet. Lets see what the community decide but please all review the linked meta (and consider voting on the answers you agree/disagree with if it's the first time you've seen it).

The question of on/off topic hinges on the distinction of whether this is an "aircraft". Technically that is something which gains lift from the air - this qualifies. Others distinguish that it must be piloted either directly or remotely, a bullet does not qualify. Is a reentry vehicle "piloted"? Hence my assertion that this is borderline.

This is basically what the Falcon 9 booster does, except it has both reaction control thrusters as well as active aerodynamic control surfaces to control attitude, so it doesn't need to be aerodynamically stable.

Define "stable lifting body". It is not a real concept in generic terms (speed matter, with enough speed you cannot be "stable"). Many reentry capsule (so bullet like shape) have some aerodynamic lift, but AFAIK none "stable" (which I think it is impossible on such speeds: just think about it: low weight, but high speed, so passing different layers of atmosphere at very high speed, so high forces (higher then weight), so unstable).

You'll most probably get a better answer on space.stackexchange

@sophit I doubt it. I came here from r the aerodynamic experts

Instinctively I would answer no.  That's why rifle barrels have rifling, and why footballs are thrown with a spiral.  Because there is no aerodynamic means to provide stability, so they must be stabilized gyroscopically.  And instinctively, if the CG was forward it would stabilize like a dart.  Aerodynamically seeking the path of least resistance.  Because what property of a forward CG might produce a positive AOA?  Conceptually it doesn't seem worthy of serious consideration.

P.S. If there are phenomena in the hypersonic/re-entry realm that are over my head and make this plausible, then please improve the question and I will delete the comment above.  Otherwise I'd be happy to expand it into a proper answer.

Really sorry that a reference to a proven design was downvoted and deleted.  Anything to do with basic fundamentals of aviation seems to have some our site members fleeing in horror.  Short answer, look at an arrow.  Now try to make it a glider.

@RobertDiGiovanni, I don't totally agree with deleting your answer, but it was really an act of mercy since you didn't answer the question.  It was a reasonable comment, that's all.

@MichaelHall fundamental research (gaining full understanding) involves building a proper foundation of facts.  Have a look a the Pyramids.  Now look at Peter Kampf's answer.

You continue to display a remarkable ability to deny and deflect.  You posed a few relevant counter question to provoke thought, (a technique I like) but again, you didn’t answer the question.  Hints are not answers.  They just aren’t.

@MichaelHall sometimes I try to steer the OP in the right direction, but your advice is well taken, and I will work on being more concise.

Definitely No!
Your rocket qualifies as a slender body. Slender body theory says that the center of pressure is near the tip, regardless of Mach number. This means that a small side force caused by a nonzero angle of attack or sideslip will amplify this angle. In other words, the rocket will be unstable in pitch and yaw. Catastrophically so if you are in a re-entry.
Most lift will occur on the nosecone and the rear body will contribute almost none. A finless rocket cannot be stabilized at any angle of attack but will tumble uncontrollably.

Won't the CP move back as the AoA rises, until it reaches the CG?

So, now we add stabilizer fins to help control these tendencies, or keep some instability a for better manuvering.

@Abdullah No, it will stay where it is until you are almost orthogonal to the flow. Remember what happens in a deep spin: Only the airplane nose will experience significant aerodynamic forces (apart from simple drag), the rest will have fully separated flow. The final posture will be orthogonal to the flow and spinning, after lots of tumbling around. Unless it breaks apart during the tumbling phase, which is very likely.

Why would we care about flow separation at those speeds?

@Abdullah We care about what happens at the opposite end of flow separation: Massive heat and aerodynamic loads over the full surface of the rocket. And the aero loads are purely drag. I understand your question that you care about lift.

How can that huge body produce no lift?

@Abdullah Please look into slender body theory. Basically, lift is only added where the width of the body increases. All what is coming behind does not contribute because the flow bending has happened already.

The Wikipedia article says it is for Stokes flow in very low Re numbers. I am at a loss as to how that has any relevance in hypersonic flow.

@Abdullah There is no Mach influence in slender body theory.

But the high speed will lead to very high Re numbers?

@PeterKämpf back in my skydiving days I experimented with trying to hold myself inverted with my body straight, head at the bottom. Terminal velocity went way up, probably to about 150kt, and in no time I would be gyrating in a kind of conical pattern that was impossible to control.  Going back to a flat arch position, the deceleration back to normal terminal velocity, the sensation was almost like falling into a swimming pool.

@Abdullah Also, viscosity is not part of slender body theory. Thus, no Re influence.

So Wikipedia lied

Am I understanding this correctly? The air thrown outward at the very tip of the nose blocks airflow to everything behind, causing the rest of the vehicle to be in a vacuum

@AbdullahisnotanAmalekite No, no air is thrown outward and there is no vacuum. But equal pressure on top and bottom except for the stations where the body widens.


text:
Is loss of engine in single engine part 23 aircraft leading to a forced landing categorized as CATASTROPHIC or HAZARDOUS?
For part 23 aircraft (CS-23 or FAR-23), there are some aircraft categories such as Normal 1,2,3 and Commuter. For instance, Cessna Grand Caravan is certified as normal category part 23.
But, it has only one engine, it means if the engine fail (complete loss of thrust), then the aircraft cannot perform continue safe flight and landing (CSFL). It could only perform a forced landing. I have some questions regarding that

Based on part 23, is a forced landing categorized as CATASTROPHIC or only HAZARDOUS?
If it is CATASTROPHIC, it means the probability of loss of engine shall be less than 1E-8 or 1E-7 and not caused by a single failure. Since it is a single engine aircraft, how can it be certified then? is the requirement for the whole powerplant or just powerplant control part?
If it is not CATASTROPHIC, is there any constraint of part 23 aircraft to fly above congested or populated area? Since a forced landing may cause a fatality if it happens in the congested area.

Edit: To summarize and clarify the questions, is total loss of engine in single engine part 23 aircraft (normal and commuter) CATASTROPHIC or HAZARDOUS?

I'd never really thought about it, but I assumed that those risk categories only applied to FAR25 Transport Category cert.

The term CFIT usually has the connotation of unexpected  & avoidable impact with terrain, while a forced landing due to engine failure is neither avoidable nor, for the time between the failure & the landing, unforeseen.  While a forced landing meets the individual words "controlled" and "flight" and "terrain" (though the ideal would be onto, rather than into), I expect that such an occurrence would be considered distinct from a classic CFIT incident & not categorized as such.

14 CFR 32.2000 states "Continued safe flight and landing means an airplane is capable of continued controlled flight and landing, possibly using emergency procedures, without requiring exceptional pilot skill or strength. Upon landing, some airplane damage may occur as a result of a failure condition.". So a failed engine is actually CSFL.

While @foot focused on the CFIT criteria, let also look at the categories you cite: "hazardous" and "catastrophic".
Definitions are found in FAR 25.4:

Hazardous failure condition means a failure condition that would reduce the capability of the airplane or the ability of the flightcrew to cope with adverse operating conditions, to the extent that there would be (i) A large reduction in safety margins or functional capabilities, (ii) Physical distress or excessive workload such that the flightcrew cannot be relied upon to perform their tasks accurately or completely, or (iii) Serious or fatal injuries to a relatively small number of persons other than the flightcrew.

Catastrophic failure condition means a failure condition that would result in multiple fatalities, usually with the loss of the airplane.

These categories are not related to CFIT events, they are used for risk management, the principle being if some failure leads to catastrophic consequences, then the failure rate must be extremely improbable, while if the consequences are only minor (another category) a frequent failure rate is acceptable. This principle is the base of risk management and mitigation in any field, and each combination can be associated with a low/medium/high/exceptional risk rating (example).
Overall the categories+likelihood combinations defines two regions:

Source.
To be airworthy, an aircraft must have its parts and systems in the acceptable region, as described in FAR 25.1309 (b):

Each catastrophic failure condition (i) Must be extremely improbable; and (ii) Must not result from a single failure.

Each hazardous failure condition must be extremely remote.

and (e):

The applicant must establish certification maintenance requirements as necessary to prevent the development of the failure conditions described in paragraph (b) of this section.

Answer to your comment

But for a single engine aircraft, is it CAT for having a forced emergency landing due to total loss of engine? If yes, than how to comply with the part 23 requirement for CAT?

The categories do not relate to the cause of a failure, so an engine failure cannot be catastrophic, but it can lead to catastrophic consequences. If a single cause can lead to catastrophic consequences, it's an unacceptable combination, the aircraft cannot be certified.
Single-engine aircraft are designed to be able to glide and land without their engine, the consequences are not catastrophic, per the definition above. If this risk occurs, the crew may have a hard time, but that's almost always manageable, and crews are specifically trained for this occurrence (e.g. when in VMC, they are trained to continuously identify safe places to land, in case an emergency landing should be required).

Hi thank you for the answer. Based on part 23 requirement, no single failure shall lead to CATASTROPHIC (CAT). But for a single engine aircraft, is it CAT for having a forced emergency landing due to total loss of engine? If yes, than how to comply with the part 23 requirement for CAT?

Thanks! By "landing without engine" it means having a forced landing outside of runway right?

@ZahiAzmi: Not necessarily, but the height above the ground will limit its gliding distance, so a forced landing in the countryside is usual (and possible, @ foot gave an extended rationale about emergency landings).

14 CFR 23.2000 states: "Continued safe flight and landing means an airplane is capable of continued controlled flight and landing, possibly using emergency procedures, without requiring exceptional pilot skill or strength. Upon landing, some airplane damage may occur as a result of a failure condition." Also Part 25 does not apply.

Isn't there a phase during the initial climb after takeoff (or even still on the ground, before take-off, after what is, I believe, called V1) where single-engine loss (single reason) would be catastrophic because a controlled stop or emergency landing is impossible during that phase so that a crash is assured?

@Peter-ReinstateMonica: In the balanced field takeoff, if the aircraft reaches V1 with all engines it can take off, accelerate, reach V2 and fly safely with one engine inoperative. Of course balanced field takeoff doesn't apply to single-engine aircraft. If the engine failure occurs before V1, the aircraft can safely stop on the runway.

@Peter-ReinstateMonica: Not really. One could set up a scenario where it would be that way, but in ordinary operation, the takeoff briefing is something like "If fails before takeoff, stop. If fails below XXX altitude (typically a few hundred feet), land on remaining runway. Between XXX and YYY altitude, choose best spot within 30 degrees of straight ahead and land there. Above yyy altitude (typically 1000 feet), return to airport. There do exist airports where that middle bit is a problem, but it is not guaranteed death, unlike the wings falling off.

The FAA defines CFIT as:

an unintentional collision with terrain (the ground, a mountain, a body of
water, or an obstacle) while an aircraft is under positive control.

An emergency landing is not CFIT, as it would not be unintentional. Part 23 also defines CSFL as:

Continued safe flight and landing   means an airplane is capable of continued controlled flight and landing, possibly using emergency procedures, without requiring exceptional pilot skill or strength. Upon landing, some airplane damage may occur as a result of a failure condition.

Since performing an emergency landing is a required skill to get a pilot license, I don't think that would be considered as "requiring exceptional pilot skill."
Additionally, part 91 defines minimum altitudes for operation (except when necessary for takeoff or landing), including a general requirement:

Anywhere.  An altitude allowing, if a power unit fails, an emergency landing without undue hazard to persons or property on the surface.

So it falls upon the pilot to not operate where an emergency landing would not be possible.

Thank you for your wonderful answer. So based on the CSFL definition, controlled emergency landing due to total loss of thrust is still counted as CSFL, is my understanding correct? Is it also applicable for Commuter category or just Normal category?

@ZahiAzmi, no, it is not CSFL because it is neither continued, nor safe.  Engine failure of a single engine airplane will result in a forced landing.  You will not be able to maintain altitude and continue to your destination as you could in a twin with an engine failure.

@fooot, this is a great explanation to correct the OP's misunderstanding about CFIT.  However, it doesn't answer any of the three numbered questions in the original post.  For that reason it is really just an excellent comment to improve the question.

"an emergency landing without undue hazard to persons or property on the surface". This reg does not require the pilot to operate at an altitude that guarantees the plane can be landed safely for the people onboard- only for people on the ground.


text:
Why are primary and supportive instruments in instrument flying different depending on the phase of flight?
For example, in straight-and-level flight, the power's primary instrument is the Air Speed Indicator (ASI), and the supportive are the engine instruments.
In another example, straight constant-airspeed climb, the primary are engine instruments, and the supportive is the ASI.
Why does the primary instrument change?

I don't normally edit other people's words, but the "change all the time" phraseology was bothering me a little...  Please review my edits and make sure they are consistent with the original intent of the question.

The whole point of classifying primary or secondary instrument (or the other method control-performance) is to teach pilots to look at what is important during different phases of flight.
When you read the Instrument Flying Handbook, it will tell you what instruments are primary and what instruments are secondary.
For example, in straight and level flight, the altimeter is primary for pitch, the airspeed is primary for power, and the turn and bank indicator is primary for bank.  If you examine what these instruments are doing at this exact moment, none of those instruments should be moving... they are constant.  Hence, the primary characteristic of primary instruments are they are constant.
Secondary instruments for straight and level flight are the VSI for pitch, power for airspeed and attitude indicator for bank.  These instruments show you how to control the airplane to return to the aircraft flight attitude you had before.  For example, you notice the airspeed start slowing.  You take a look at the secondary instrument, power and notice that it has slowly backed off.  Increasing power should then cause the airspeed to return to normal.  Hence, the primary characteristic of secondary instruments are they show trends.
Reading the manual, you will also discover that anytime you change the pitch or bank of the airplane you should look at the attitude indicator.  It is primary instrument for all changes in the flight attitude of the airplane.  For example, if you want to change between straight and level to a constant airspeed climb, you will use the attitude indicator to set the pitch attitude and then verify the airspeed is climbing at the correct speed.

Because during certain maneuvers, certain instruments provide more relevant information to a pilot than do others.
In straight and level flight, the heading indicator provides more relevant bank information as it tells you whether your current bank attitude is maintaining the desired heading.  The altimeter is providing the most relevant pitch information as it’s telling you whether your current pitch will maintain the correct altitude.  The airspeed indicator is the best indication for your required power as it’s telling you if the current power settings will maintain the desired airspeed.  Other instruments provide supporting information.  Here in straight and level flight the attitude indicator provides good supporting bank and pitch indications, the VSI provides supporting pitch information, etc.  Supporting instruments change to primary instruments in certain maneuvers and vice versa.
And, incidentally, in a established constant airspeed climb, your primary for bank is the heading indicator and your primary for pitch is the airspeed indicator.  Primary for power is either the tachometer, manifold pressure, torque, or fan speed gauges, depending on the aircraft you are flying.  Typically you would set to maximum continuous power in a light airplane for this situation.

The answers already provided are very good, this answer is meant to say the same thing in different terms, not to contradict what has already been written.
A primary instrument is one that provides a value for you to hold. If you want to maintain 5,000 feet, then the altimeter is the primary instrument because though there are other instruments that can tell you about your altitude, the altimeter is the only one that can show you what your altitude is.
Similarly, when entering a climb, the attitude indicator is primary for pitch, because you should be setting a particular pitch and the attitude indicator is where you look to see what degree of pitch you have.
Primary instruments give you a target, supporting instruments help you hit it.

All the answer except for quiet flyer's are spot on. Although, I do appreciate, acknowledge and understand his point of view. 90 years of instrument flight has dictated which instrument is primary and which is secondary for each state of flight. It is not arbitrary. The way to look at it is that you are trying to achieve a certain outcome without the benefit of your usual feedback. You can not rely on on sight, sound (in some cases), touch (seat of the pants), or sense of motion and balance.
Say for instance, if you are flying IMC. And, you want to fly straight and level in unaccelerated flight. The Airspeed Indicator is your primary for power because (hyperbolically) who cares how fast your prop is turning. Your RPMs are not an outcome. At least, it is not your desired outcome. Your desired outcome is a constant airspeed. Your engine performance is tangential to that. In fact, your RPMs are going to change as your Density Altitude changes even at the same airspeed. Then, throw into the mix a constant speed or otherwise adjustable prop, your Tach becomes almost unusable as a primary for power except when you are changing from one maneuver state to another.
Personally, I think that the Control and Performance Method is much better at addressing all of the posters' views. Especially quiet flyer's. In reality, everyone that I fly with uses Control and Performance intuitively. In VMC, Control and Performance is automatic and second nature because you are using sight, sound, and sense of motion primarily to fly the plane. It can still be used when those senses are not available in IMC.
Where the Primary and Supporting Method shines is when you start losing the use of your instruments in partial panel flight. It tells and teaches you what instruments (supporting instruments) can be used when the ones you want (primary instruments) are not available or inop.

Good answer Dean.  For both methods, the attitude indicator is primary whenever the flight attitude (pitch and roll) is changed.  I don't think the PRI/SEC method guys fully understand that.  It is in the IFH.  I was taught the PRI/SEC method but now teach a combination of the two.

"Why can't we use just one as a primary instrument?" -- sometimes we can.  I've seen video of a Cessna 120 being flown in cloud with all gyro instruments covered up.  The pilot's hands were not touching the control stick or throttle.  The airspeed, tachometer, altimeter, and vertical speed indicator (if present) were not covered up, but they might as well have been.  The aircraft was trimmed for a controlled descent in fairly smooth air and emerged from the bottom of cloud layer about 5 minutes after the video started.  The magnetic compass was obviously being used as the primary instrument.  Steering corrections were accomplished by using the rudder pedals to bank the plane slightly via the "dihedal effect" created primarily by the aircraft's high-wing configuration. Don't try this at home unless you understand what heading to use, and why.
This whole business of classifying instruments into "primary" and "supporting" is rather subjective and arbitrary.  Who gets to decide that the "in straight-and-level flight, the power's primary instrument is ASI", rather than, say, the tachometer?  Maybe a lower-than-expected airspeed in level flight is caused by ice on the wings, or by the pilot forgetting to raise the landing gear, rather than by poor engine performance.

delete "the" from

I can appreciate your point of view. Flying partial panel in actual IMC is not something you want to do. It is neither easy nor desirable. If you lose any instrument in IMC, it is cause for declaring an emergency. As for the video you saw, was the rate-of-turn or turn-bank coordinator covered (if they had an electric one). Considering the errors inherent in a mag comp, the ROT would be a better substitute primary instrument to stay straight in a constant descent than a compass.

As far as the airspeed scenario... If you have ice on your wings or stuck landing gear, you can forget about the Tach. The Airspeed Indicator is key. You will have to use pitch and throttle in any combination necessary to ensure your angle of attack and airspeed is sufficient to provide enough lift. At the same time, you need to keep from stalling while recognizing your speed limitations. You can use your ears to judge your engine management and your temp and pressure gauges to judge engine health. In your scenario, your Tach becomes unreliable for what's most important.

@DeanF -- But the pitot tube may ice up as well.

More planes have pitot heat than leading edge boots. You have an out. The key is being able to recognize the issue. Although, you should have had the pitot heat on in the first place. That being said, your example is why Primary-Supporting is still useful. You will immediately and without hesitation know what to use for Power control and management if your airspeed indicator goes completely inop. Airspeed is still king. So is its indicator when it is working.


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What happens if the pitot tube freezes?
Can somebody explain this a bit for me?
I.e. as I understand:
If the Pitot tube freezes/ becomes blocked, the ASI would no longer have a change in the pressure (dynamic + static) from the pitot tube, but would have only change in static pressure from the static port.
So, in that case:
If on ground (before takeoff), the ASI would not become active.
A. On climb, the total pressure from the pitot remains the same, whereas the static pressure decreases, so relatively the pitot pressure would be higher compared to the previous moments, there would be some change, but the ASI would under read. Is that correct?
B. If climbing in cruise, same as A. above, correct?
C. On descent, the static pressure would increase, but the total pressure (dynamic-and-static-pressure) from the pitot tube would be very low, so the ASI would under-read, is that correct?
What about with the static port? I.e. if the static port is freezed over?
D. In a climb, there is no change in the static pressure, and the total pressure systems from the pitot is working, the ASI would still under-read due to the density decrease in altitude. Is that correct?
E. In a descent, the static port is frozen, no change in static pressure, but the lowering altitude would mean higher density, therefore the ASI would over-read, is that correct?
Any help would be greatly appreciated.

Airspeed = Pitot Pressure (Dynamic + static) - Static Pressure (static)

For a pitot tube that freezes (along with the drain) but the static port is clear.

As aircraft altitude increases, the static pressure decreases.  Since the pitot pressure remains the same, the difference between pitot and static pressure increases and the airspeed indicator will read higher than it should.
As aircraft altitude decreases, the static pressure increases.  The difference between the pitot and static pressure decreases and therefore the airspeed indicator reads lower than it should.

A. If on ground (before takeoff), the ASI would not become active.

The airspeed indicator will increase with altitude as stated above. If the PITOT STATIC differential is 1.0" Hg, the airspeed indicator would read 145 KTS.  If you were to takeoff with a clogged pitot port (and drain), and climb 1000' MSL, the airspeed indicator would also read 145 KTS.
Luiz Monteiro Pitot-Static Simulator

If the static port is clogged while the pitot port is clear, you can use the same formula above to determine the result.

If the aircraft climbs with a clogged static port, the pressure differential will decrease (PITOT pressure decreasing while static pressure remains the same) and the airspeed will read lower than it should.
If the aircraft descends, the pitot pressure increases while the static pressure remains the same and the differential increases, causing the airspeed to read higher than it should.


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Why don’t ionocrafts (ionic propelled aircrafts) use ducted ionic thrusters instead of open air thrusters?
The most famous example of inonocraft is the aircraft with no-moving parts propulsion made by the MIT.

A problem of ionic thrusters is that they only generate a gram of thrust for every watt, which is really inneficient.
However, if one would to make a ducted ionic thruster that worked like a ducted turbine, wouldn't the thrust increase?
The closest thing I could find was this video that went viral some time ago showing an attempt of making a multistage ionic thruster.

The other ionic thruster that is worth talking about is an atmospheric ionic thruster satellite that will take the air from the edge of space and accelerate it. But that is an special case, since the aircraft will be sent there by a rocket and maintain orbit on its own. It won't take off on its own.

My patented Self-Contained Ion Powered (propelled) Aircrafts and Ion Propelled Vehicle, produce about 7 grams of thrust per watt at the batteries and roughly 14 grams of thrust per watt at the airframe.  That is much more than one gram of thrust per watt.  It has flown verifiably on camera for almost two minutes at a time with onboard power. The later MIT standard glider shaped version produced 1/2 gram of thrust per watt, according to the table on the last page of their paper.  By the way Rory, please do not delete my posts!  They are entirely relevant.

A problem of ionic thrusters is that they only generate a gram of thrust for every watt, which is really inneficient.

The Wikipedia article says that about the early versions. I don't think it applies to the Ethan Krauss' self-contained prototype.

However, if one would to make a ducted ionic thruster that worked like a ducted turbine, wouldn't the thrust increase?

The efficiency wouldn't. To increase efficiency—of any aerodynamic engine—you need to increase the cross-section. This is because thrust is equal to the change in momentum of the airflow, and momentum is mass times velocity, but the change in kinetic energy of the flow is equal to ½ mass times velocity squared and the engine has to provide that energy.
Therefore it is better to have a lot of low-power cells side-by-side, to affect a lot of air, than to have them one after another in a duct, accelerating the air to a higher velocity. That's why the Ethan Krauss' prototype is a big hexagonal wire mesh grid.

This is Ethan Krauss. My VTOL Ion Propelled Vehicles have onboard power. Rather than a wire mesh, they have streamline wing shaped collectors made of concentric hexagons. Those surfaces are currently metalized with a thin coating of silver. The crafts produce many grams of thrust per watt.  Much of the airflow that propels these crafts generally speaking, comes in from the sides, so adding ducts not only reduces lift but adds weight.  Building them with multiple stages also creates less thrust per gram, unless the stages are spaced far apart from one another.

Good to have the man himself here. Welcome!

There is another one besides the MIT glider version.  Their glider flew technically in ground effect for a few seconds, after being bungee launched.   The much earlier Ion Propelled Vehicle was patented for lifting its power supply with ion propulsion.  Usually tested flying in VTOL mode, for up to minutes at a time.  It has also flown on its side gaining aerodynamic lift.  It is less stable than a standard glider shape since it consists of concentric hexagon shaped collector/wings, but it was conceived of from the start to be an efficient ion propelled VTOL vehicle and less like a glider.

Is it correct to ask about "inonocrafts" (the plural?) Are there any other examples besides the 5lb MIT one?
The theoretical thrust gain from a duct would have to overcome the additional weight of the duct.
And this one example is just the very first basic prototype. This is sort of like asking "Why didn't the Wright Brothers use turboprops?"

The Wikipedia article links two examples. And the other one is a vertical take-off one.

I am not sure there even is any theoretical gain from a duct here. Ducted fans work because the fan blades benefit from an end-plate, but the electric field does not have any edge that could get similar benefit.

I discounted that one, because it couldn't lift its own power supply. Like an electric car prototype, but you need a really long extension cord.

Does it? The article says “The first ion-propelled aircraft to take off and fly using its own onboard power supply was a VTOL craft developed by Ethan Krauss of Electron Air in 2006.”.

Ah, I was reading/conflating the wrong one.

Tethered ionocraft are a perfectly good testbed for trying out techniques for improving efficiency.


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Do critical angles of attack vary from plane to plane? Can a plane have higher or lower critical AoA than the other?
I have done a lot of research and I have found out that different planes have different climb angles. Some look like they are climbing at a higher rate than others. Why is this? Do some aircraft have higher or lower critical AoA than the other? If yes, then why? Is it the design of the airfoil?

I guess it also depends on the kind of engine. E.g. turbojets may have a higher critical aoa than propeller planes.

I think you're misunderstanding the concept of Angle of Attack. The Angle of Attack is the angle between the plane's flight path through the air and the airfoil's chord.

(Source: https://www.apstraining.com/resource/three-critical-angles/)
If an airplane is pitched up 12° from the horizon, but climbing at 10°, then its angle of attack is only 2°.
The angle of attack doesn't have anything to do with climb performance. The maximum rate of climb and angle of climb are (mostly) determined by the amount of thrust the engine is creating, the efficiency of the wing, and the weight of the plane.
Different airfoil designs will indeed have different critical AoAs.

To make a minor but important clarification, Flight path must me measured relative to the airmass, not the ground, of course (this is why the parenthetical note says assumes no wind -  Flight path markers on attitude indicators are almost always depicted relative to the ground) . This is important, especially as any reference to angle of climb can potentially confuse the issue, as generally that is measured relative to the ground, not the airmass.

@CharlesBretana I've edited it slightly.


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Swept outer section of the wing on turboprops
It's quite common to see turboprops having a wing with the outer part of the wing with a small sweep (around 3 deg at 1/4 of the chord). What's the reason?
It's usually stated that wing taper is used mainly for reducing the critical mach number or to fix the CG problem. For turboprops such as ATR-42/72 or Dash-8 the effect of Mach number is still negligible and it's unlikely that both of these planes and their subvariants have this feature because they needed to fix the CG.

If the main spar is straight across, the wing isn't swept, just tapered.

Don't think of those wings as swept.  As far as aerodynamics are concerned, they are straight wings.
Think of them as tapered -- The taper is there to improve the lift distribution.
The line about which the wing has zero sweep is chosen for non-aerodynamic purposes (since there is little aerodynamic benefit or penalty to one choice or another).
In these cases, the wing is un-swept at the mid-chord or possibly the 3c/4 line.  This is usually for simplicity of layout of structures and systems.
In a related manner, some aircraft twist their wings about the line that the control surfaces are attached to.  This leads to a straight and true aft spar that is easy to hang control surfaces from.

It's quite common to see turboprops having a wing with the outer part of the wing with a small sweep

At least for the ATR-42/72/Dash-8, the wing possesses no sweep as it can be gathered from the following cutaway (source):

As it can be seen, the wing spars (185 in that picture) are basically straight and perpendicular to the fuselage. An angle different from 90° would denote a swept wing, which is not the case here.
What you're referring to is actually the taper ratio i.e. the fact that the chord length reduces spanwise from root to tip.

What's the reason?

This is done to reduce lift (which is proportional to the chord length) from root to tip in order to optimise its spanwise distribution, reduce the relevant induced drag and optimise the wing structure, as explained in several answers here, like for example this one.

It's usually stated that wing taper is used mainly for reducing the critical mach number or to fix the CG problem

Tapering the wing doesn't really help in reducing the critical Mach number: sweeping the wing and choosing an appropriate airfoil (so called supercritical airfoil) reduces transonic effect with the relevant drag increase. If you find a CG problem after the design freeze then changing the taper and/or the sweep angle is definitely not a quick way to fix this problem since it entails a major (and costly) redesign of the wing itself.

There are two 185's in the diagram, both the forward and aft outboard spars are labeled the same.  Which of these is the 'main' spar is certainly up for debate.  The aft spar is un-swept in this case and it lies at about 60% chord.
Notice that both the flaps and the ailerons are attached to the aft spar.  In this case, it is evident that the aft spar was placed to coincide with the control surface hinge line -- and that said line was set to be un-swept.

@RobMcDonald: right, corrected, thanks for the comment


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Examples of jet turbine values for combustions, fuel flow, Q of fuel, temp of exhaust, etc. for classroom use
I am trying to build jet turbine examples for a class and I want to use actual values for the mass flow of fuel and air, combustion temps, exhaust temps, exhaust composition (CO, CO2,H20,o2, NOx) ext…
I thought this would be a quick 20 minutes of googling but I cant seem to find a clean consolidated source for the information.
I want to have the students run multiple scenarios showing a lean mix lowers the adiabatic flame temp and produced less NOx vs older designs that ran rich and had higher emissions rates. The values don’t need to be 100% accurate but in the ballpark would be nice. Links to sources would be best. Generic values used commonly in industry as back of a napkin values are a start.
I see that there are a lot of questions in the same ballpark of this but I (from my first run through) did not see one with all the information. Sorry if this is a repeat there are 1000’s of Questions on here to sift through.

For this answer I found a nice reference that had quite a bit of data, real and simulated, maybe it has something for you?

Thanks for the quick response, I'm looking more for combustion related information but the temperature data maybe useful to compare once I have built a working model.

You could get an R/C jet as a working model.  You would not want a homebuilt exploding in class.  Diagrams should suffice.  A field trip to the airport might be fun.

Funny enough the only groups that share all the data for their engines are model jets. Being models they don't have quite as much of the technical data but still they are open. I found a model jet used for education that has a lot of accompanying data.
https://www.turbinetechnologies.com/educational-lab-products/turbojet-engine-lab

Old data is still data.
Aircraft Emissions Characterization: F101 and F110 Engines
March 1990, Chester SpicerM. W. HoldrenD. L. Smith, et al
1990's report on an engine design that, according to wiki, was the basis for the CFM 56.
https://www.researchgate.net/publication/235111468_Aircraft_Emissions_Characterization_F101_and_F110_Engines
Has ambient and combustion temperature data, ppm of NOX, NO2, CO, CO2, thrust output, mass flow of both air and fuel. Really the only thing it is missing for a full combustion analysis and ideal jet propulsion analysis is the efficiencies.
I just built the model assuming everything was ideal and needed an HV of 55,000 (which is way above jet fuel) to make it match the data.


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Are the F-18 movements in the Top Gun: Maverick movie really possible?
Lately, I watched the new movie Top Gun: Maverick. A few times Maverick made some impressive movements when he was getting hunted by another fighter jet. He somehow goes slightly up, brakes and gets behind the jet.
I tried to find a look-a-like image to show the movement. Here you can see a screenshot of the trailer on YouTube, which shows the movement of getting behind a fighter jet:

So I was wondering if it is really possible to get behind a fighter jet like this way?

It is called Cobra Manoeuvre and can be performed for sure by some fighters (Saab Draken for example).

Yes, it's the cobra. Many fighters can do it but among F-18s only one modified aircraft is capable (NASA F-18 High Alpha Research Vehicle (HARV)).

@user3528438 I think some kind of AoA protection prevents most fighters from being able to perform this manoeuvre.  The diversion manoeuvre of the second fifth-gen fighter looks like a very tight Herbst manoeuvre instead (?)

Similar to 'VIFFing' in the Harrier. Looks great, practically no combat application.

The "Cobra maneuver" is primarily an airshow stunt. To my knowledge it has never been performed or verified in actual combat.

At least not successfully, because it presents a huge target for the other plane’s guns.

@StephenS Assuming the other pilot even has guns.


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What is the metal plating on the back of the A-6 for?
What is the perforated metal plating on the back of the A-6/EA-6A for?

Source: Wikipedia

You mean on the belly to the aft, under the MAR of MARINES?

It is a speed brake, a plate hinged at the forward edge which can be hydraulically extended to increase aerodynamic drag. It had been used on early models, and the speed brakes were mostly replaced by firmly riveted plates later, because flight test revealed handling problems. The wake of the speed brake disturbed the flow around the horizontal tail especially during approach and landing. They were replaced by wing tip brakes, and from the 304th aircraft on were completely omitted.
See here for further details.
Below is a picture of an EA-6B with the wing tip speed brakes (circled red) extended:

(Source: navy.mil)

Now I'm curious--are those wingtip speed brakes, or a split aileron such as on an A-10 (http://cdn-www.airliners.net/aviation-photos/photos/6/2/3/1681326.jpg)?

I never realized those aircraft have (at least) 2 types of speed brakes (the one highlighted in the question and the one on the wingtip)

@Marius, those are wingtip speedbrakes.  They have no other function like a split aileron might.

P.S. I edited this answer to correctly identify the plane in the photo as an EA-6B. The 2 seat EA-6A pictured in the question had these fuselage speedbrakes because it was an A-6 airframe, the 4 seat EA-6B pictured here did not. FWIW, the first prototype EA-6B was built from an A-6 airframe and had these as well, but production models, (while based on the A-6 and similar in appearance) were actually a very different airframe.  (Also, I had to check my log book, but I have 21 flights in 160709.  Had a couple landing gear problems with it...)

@MichaelHall Thank you, your help is very much appreciated.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


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Does tip dihedral have lesser increase in lift over a full dihedral wing?
I have read some articles mentioning that a full dihedral reduces the lift produced by a wing. So what about a tip (or cranked) dihedral that has the angled part starting from the mid-wing rather than the root of the wing?

What does your analysis tell you?

That resembles something they did with the Phantom jet.  But look at its tail.

The ratio of vertical lift to total lift in a bank angle is cosine bank angle.  Cosine of 5 degrees is 0.996.  Dihedral "banks" each wing very slightly from vertical.  Typical dihedral angles affect lift very little.
Working this problem a bit further, we find that the cosine function is non-linear as angle increases.  This means that half a wing at a steeper angle with half a zero angle actually has less vertical lift than a full wing dihedral of the same fuselage to wing tip angle.
A "cranked dihedral" would also be more complicated to build and not as strong.  The design is sometimes seen in older model gliders. Some full scale aircraft may have winglets, but are plenty stable without extra dihedralling at the wingtips.

It's not harder to build at all.  You have to put the joint for the change in angle in there somewhere. The Thorp T-18 did it to allow each wing to be single 4 ft skin wraps, with a straight spar carrythrough and the joint at mid span.

Yes, but the "joint" now has a sharp angle.  I'd rather brace the whole dihedral angle with a cross beam inside the fuselage.  The tips do provide a bit more ground clearance.

The angle is unimportant.  You are creating a splice at the change in angle, and putting it at mid span means the  splice can be much lighter, the stresses being lower, and you now have a straight one piece spar carrythrough through the fuselage without any splices at the highest stress location.  Most designers do it at the wing root because the wing is designed to be removeable so the connection is done at the root and might as well put the dihedral break there as well.

I can see along an entire rib structure a splice is feasible, but if anything I would favor a bit of anhedral (from a dihedralled wing).  Birds such as seagulls and geese do this naturally to improve crosswind control.  But it is a workable design.

Minor observation--if the design objective definitely includes a wing that can be disassembled into at least two pieces, much weight can be saved by making a one-piece center section with removable tips, as opposed to a wing that breaks apart right at the center section.  (I.e. don't put a junction at the point where the bending load is greatest.)  In such a case, if the tips are removable anyway, there may not be a significant weight (or roll inertia) penalty incurred by designing a dihedral break coincident with the point where the removable tip joins the wing center-section.  (ctd)

(...ctd)  Case in point-- "Gentle Lady" r.c. sailplane-- plans included options for one-piece wing or three-piece wing.   In either case, there was also a dihedral break at the centerline, so the entire wing shape could be described as "polyhedral"-- 4 panels total, 2 on each side, each with its own unique orientation.  Completed shape of the assembled wing was the same regardless of whether the 1-piece or 3-piece option was chosen.

@quietflyer I was thinking similarly, and models generally are proportionally lighter.  Full scale, a flange would have to be made to support the "break" in each wing angle.  A center angle support could be built into the fuselage.  Less roll inertia to overcome, ergo, smaller ailerons.  Just doesn't pass my sniffer, especially with the T tail.  Happy 4th.

The main performance penalty incurred by dihedral is arguably not the slight loss of "verticality" in the lift vector generated by each wing, but rather the fact that if any accidental sideslip is present, the angle-of-attack of the "upwind" wing will be increased and the angle-of-attack of the "downwind" wing will be decreased.  (That's how dihedral "works" to create roll stability, after all.)  Even disregarding the additional drag incurred by whatever aileron input is needed to cancel this roll torque, due to the shape of the curves of the lift and drag coefficients (versus angle-of-attack), flying around with the two wings at different angles-of-attack is generally less efficient than flying around with the two wings at the same angle-of-attack.
Making the wing flat in the middle and reserving the dihedral for the tips-- where it will have the most benefit in terms of maximizing the stabilizing roll torque generated by a sideslip-- minimizes this drag penalty during accidental (and perhaps transient) sideslips, for any given desired degree of overall "effective dihedral", i.e. any desired amount of roll torque generated by a given sideslip angle.


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What determines military aviation squadron numbers?
What determines the numbers and the names of squadrons, airwings, and other military aviation units? Does it go in order of when they were established?

Many factors are involved, that create little rhyme or reason.  Much of the influence is historical.  If the Army signal corps 123 battalion was transformed to become an Army Air Corps Battalion in, say, WWI, then that might also later become the 123 Fighter Wing years later.  Also, the Army went through a period in WWII where there were categories that drove numbers that were thus out of sequence.  For instance, for a time, airborne units were all given 5xx designations (like the 509th PIR) even if the sequential numbers were only up to say 300.  So it's a giant "it depends" non-answer.

Tradition. History. Phase of the moon. How hungover the general was that day. Who knows. :)